319204
Zinc reacts with \({\rm{CuS}}{{\rm{O}}_{\rm{4}}}\) according to the equation \(Zn + CuS{O_4} \to ZnS{O_4} + Cu\). If excess of zinc is added to 100ml, 0.05M \({\rm{CuS}}{{\rm{O}}_{\rm{4}}}\), the amount of copper formed in moles will be
1 5
2 0.5
3 0.05
4 0.005
Explanation:
1 mole of \({\rm{CuS}}{{\rm{O}}_{\rm{4}}}\) gives 1 mole of Cu \({\rm{0}}{\rm{.1L \times 0}}{\rm{.05M = 5 \times 1}}{{\rm{0}}^{{\rm{ - 3}}}}\) mole of \({\rm{CuS}}{{\rm{O}}_{\rm{4}}}\) gives \({\rm{5 \times 1}}{{\rm{0}}^{{\rm{ - 3}}}}\) mole of Cu.
CHXII02:SOLUTIONS
319205
What volume of 0.8M solution contains 0.1 mole of the solute?
319206
The density of a solution prepared by dissolving \(120 \mathrm{~g}\) of urea (molecular mass \(=60 \mathrm{~g} \cdot \mathrm{mol}^{-1}\) ) in \(1000 \mathrm{~g}\) of water is \(1.15 \mathrm{~g} / \mathrm{ml}\). The molarity of this solution is:
1 \(1.78 \mathrm{M}\)
2 \(1.02 \mathrm{M}\)
3 \(2.05 \mathrm{M}\)
4 0.50
Explanation:
Mass of solute taken \(=120 \mathrm{~g}\) Molecular mass of solute \(=60 \mathrm{~g}\) Mass of solvent \(=1000 \mathrm{~g}\) Mass of solution \(=1000+120=1120 \mathrm{~g}\) Volume of solution \(=\dfrac{\text { mass }}{\text { Density }}=\dfrac{1120}{1.15} \mathrm{ml}\) \(=973.9 \mathrm{ml}=0.9739 \mathrm{~L}\) Now, no. of moles of solute \(=\dfrac{120 \mathrm{~g}}{60 \mathrm{~g}}=2\) \(\therefore\) Molarity \(=\dfrac{\mathrm{n}}{\mathrm{V}(\mathrm{L})}=\dfrac{2}{0.9739}=2.05 \mathrm{M}\)
319204
Zinc reacts with \({\rm{CuS}}{{\rm{O}}_{\rm{4}}}\) according to the equation \(Zn + CuS{O_4} \to ZnS{O_4} + Cu\). If excess of zinc is added to 100ml, 0.05M \({\rm{CuS}}{{\rm{O}}_{\rm{4}}}\), the amount of copper formed in moles will be
1 5
2 0.5
3 0.05
4 0.005
Explanation:
1 mole of \({\rm{CuS}}{{\rm{O}}_{\rm{4}}}\) gives 1 mole of Cu \({\rm{0}}{\rm{.1L \times 0}}{\rm{.05M = 5 \times 1}}{{\rm{0}}^{{\rm{ - 3}}}}\) mole of \({\rm{CuS}}{{\rm{O}}_{\rm{4}}}\) gives \({\rm{5 \times 1}}{{\rm{0}}^{{\rm{ - 3}}}}\) mole of Cu.
CHXII02:SOLUTIONS
319205
What volume of 0.8M solution contains 0.1 mole of the solute?
319206
The density of a solution prepared by dissolving \(120 \mathrm{~g}\) of urea (molecular mass \(=60 \mathrm{~g} \cdot \mathrm{mol}^{-1}\) ) in \(1000 \mathrm{~g}\) of water is \(1.15 \mathrm{~g} / \mathrm{ml}\). The molarity of this solution is:
1 \(1.78 \mathrm{M}\)
2 \(1.02 \mathrm{M}\)
3 \(2.05 \mathrm{M}\)
4 0.50
Explanation:
Mass of solute taken \(=120 \mathrm{~g}\) Molecular mass of solute \(=60 \mathrm{~g}\) Mass of solvent \(=1000 \mathrm{~g}\) Mass of solution \(=1000+120=1120 \mathrm{~g}\) Volume of solution \(=\dfrac{\text { mass }}{\text { Density }}=\dfrac{1120}{1.15} \mathrm{ml}\) \(=973.9 \mathrm{ml}=0.9739 \mathrm{~L}\) Now, no. of moles of solute \(=\dfrac{120 \mathrm{~g}}{60 \mathrm{~g}}=2\) \(\therefore\) Molarity \(=\dfrac{\mathrm{n}}{\mathrm{V}(\mathrm{L})}=\dfrac{2}{0.9739}=2.05 \mathrm{M}\)
319204
Zinc reacts with \({\rm{CuS}}{{\rm{O}}_{\rm{4}}}\) according to the equation \(Zn + CuS{O_4} \to ZnS{O_4} + Cu\). If excess of zinc is added to 100ml, 0.05M \({\rm{CuS}}{{\rm{O}}_{\rm{4}}}\), the amount of copper formed in moles will be
1 5
2 0.5
3 0.05
4 0.005
Explanation:
1 mole of \({\rm{CuS}}{{\rm{O}}_{\rm{4}}}\) gives 1 mole of Cu \({\rm{0}}{\rm{.1L \times 0}}{\rm{.05M = 5 \times 1}}{{\rm{0}}^{{\rm{ - 3}}}}\) mole of \({\rm{CuS}}{{\rm{O}}_{\rm{4}}}\) gives \({\rm{5 \times 1}}{{\rm{0}}^{{\rm{ - 3}}}}\) mole of Cu.
CHXII02:SOLUTIONS
319205
What volume of 0.8M solution contains 0.1 mole of the solute?
319206
The density of a solution prepared by dissolving \(120 \mathrm{~g}\) of urea (molecular mass \(=60 \mathrm{~g} \cdot \mathrm{mol}^{-1}\) ) in \(1000 \mathrm{~g}\) of water is \(1.15 \mathrm{~g} / \mathrm{ml}\). The molarity of this solution is:
1 \(1.78 \mathrm{M}\)
2 \(1.02 \mathrm{M}\)
3 \(2.05 \mathrm{M}\)
4 0.50
Explanation:
Mass of solute taken \(=120 \mathrm{~g}\) Molecular mass of solute \(=60 \mathrm{~g}\) Mass of solvent \(=1000 \mathrm{~g}\) Mass of solution \(=1000+120=1120 \mathrm{~g}\) Volume of solution \(=\dfrac{\text { mass }}{\text { Density }}=\dfrac{1120}{1.15} \mathrm{ml}\) \(=973.9 \mathrm{ml}=0.9739 \mathrm{~L}\) Now, no. of moles of solute \(=\dfrac{120 \mathrm{~g}}{60 \mathrm{~g}}=2\) \(\therefore\) Molarity \(=\dfrac{\mathrm{n}}{\mathrm{V}(\mathrm{L})}=\dfrac{2}{0.9739}=2.05 \mathrm{M}\)
319204
Zinc reacts with \({\rm{CuS}}{{\rm{O}}_{\rm{4}}}\) according to the equation \(Zn + CuS{O_4} \to ZnS{O_4} + Cu\). If excess of zinc is added to 100ml, 0.05M \({\rm{CuS}}{{\rm{O}}_{\rm{4}}}\), the amount of copper formed in moles will be
1 5
2 0.5
3 0.05
4 0.005
Explanation:
1 mole of \({\rm{CuS}}{{\rm{O}}_{\rm{4}}}\) gives 1 mole of Cu \({\rm{0}}{\rm{.1L \times 0}}{\rm{.05M = 5 \times 1}}{{\rm{0}}^{{\rm{ - 3}}}}\) mole of \({\rm{CuS}}{{\rm{O}}_{\rm{4}}}\) gives \({\rm{5 \times 1}}{{\rm{0}}^{{\rm{ - 3}}}}\) mole of Cu.
CHXII02:SOLUTIONS
319205
What volume of 0.8M solution contains 0.1 mole of the solute?
319206
The density of a solution prepared by dissolving \(120 \mathrm{~g}\) of urea (molecular mass \(=60 \mathrm{~g} \cdot \mathrm{mol}^{-1}\) ) in \(1000 \mathrm{~g}\) of water is \(1.15 \mathrm{~g} / \mathrm{ml}\). The molarity of this solution is:
1 \(1.78 \mathrm{M}\)
2 \(1.02 \mathrm{M}\)
3 \(2.05 \mathrm{M}\)
4 0.50
Explanation:
Mass of solute taken \(=120 \mathrm{~g}\) Molecular mass of solute \(=60 \mathrm{~g}\) Mass of solvent \(=1000 \mathrm{~g}\) Mass of solution \(=1000+120=1120 \mathrm{~g}\) Volume of solution \(=\dfrac{\text { mass }}{\text { Density }}=\dfrac{1120}{1.15} \mathrm{ml}\) \(=973.9 \mathrm{ml}=0.9739 \mathrm{~L}\) Now, no. of moles of solute \(=\dfrac{120 \mathrm{~g}}{60 \mathrm{~g}}=2\) \(\therefore\) Molarity \(=\dfrac{\mathrm{n}}{\mathrm{V}(\mathrm{L})}=\dfrac{2}{0.9739}=2.05 \mathrm{M}\)
319204
Zinc reacts with \({\rm{CuS}}{{\rm{O}}_{\rm{4}}}\) according to the equation \(Zn + CuS{O_4} \to ZnS{O_4} + Cu\). If excess of zinc is added to 100ml, 0.05M \({\rm{CuS}}{{\rm{O}}_{\rm{4}}}\), the amount of copper formed in moles will be
1 5
2 0.5
3 0.05
4 0.005
Explanation:
1 mole of \({\rm{CuS}}{{\rm{O}}_{\rm{4}}}\) gives 1 mole of Cu \({\rm{0}}{\rm{.1L \times 0}}{\rm{.05M = 5 \times 1}}{{\rm{0}}^{{\rm{ - 3}}}}\) mole of \({\rm{CuS}}{{\rm{O}}_{\rm{4}}}\) gives \({\rm{5 \times 1}}{{\rm{0}}^{{\rm{ - 3}}}}\) mole of Cu.
CHXII02:SOLUTIONS
319205
What volume of 0.8M solution contains 0.1 mole of the solute?
319206
The density of a solution prepared by dissolving \(120 \mathrm{~g}\) of urea (molecular mass \(=60 \mathrm{~g} \cdot \mathrm{mol}^{-1}\) ) in \(1000 \mathrm{~g}\) of water is \(1.15 \mathrm{~g} / \mathrm{ml}\). The molarity of this solution is:
1 \(1.78 \mathrm{M}\)
2 \(1.02 \mathrm{M}\)
3 \(2.05 \mathrm{M}\)
4 0.50
Explanation:
Mass of solute taken \(=120 \mathrm{~g}\) Molecular mass of solute \(=60 \mathrm{~g}\) Mass of solvent \(=1000 \mathrm{~g}\) Mass of solution \(=1000+120=1120 \mathrm{~g}\) Volume of solution \(=\dfrac{\text { mass }}{\text { Density }}=\dfrac{1120}{1.15} \mathrm{ml}\) \(=973.9 \mathrm{ml}=0.9739 \mathrm{~L}\) Now, no. of moles of solute \(=\dfrac{120 \mathrm{~g}}{60 \mathrm{~g}}=2\) \(\therefore\) Molarity \(=\dfrac{\mathrm{n}}{\mathrm{V}(\mathrm{L})}=\dfrac{2}{0.9739}=2.05 \mathrm{M}\)
